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3 years ago

8

a boy runs from his house to school at speed 5m/s on a straight road and returns with speed of 10m/s.The distance between the ho

use and school is 100m. Find his average velocity.... with explaination please

1 answer:

CaHeK987 [17]3 years ago

3 0

Explanation:

time spent to run from house to school=100/5=20s

time spent to return from school=100/10=10s

average velocity=200m/(10+20)

=6.66m/s

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What is the difference between forming a hypothesis and making a prediction based on the hypothesis?

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Answer:

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3 years ago

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In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to

goldenfox [79]

Answer:

17304 J

Explanation:

Complete statement of the question is :

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .

Part A

How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Solution :

$T$ = tension force in the tow rope = 350 N

$L$ = length of the incline surface = 120 m

$W_{t}$ = work done by tension force = ?

The tension force acts parallel to incline surface, hence work done by tension force is given as

$W_{t} = T L\\W_{t} = (350) (120)\\W_{t} = 42000 J$

$h$ = height gained by the rider = 30 m

$m$ = total mass of rider and tube = 84 kg

Potential energy gained is given as

$U = mgh\\U = (84) (9.8) (30)\\U = 24696 J$

$Q$ = Thermal energy created

Using conservation of energy

$Q = W_{t} - U\\Q = 42000 - 24696\\Q = 17304 J$

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3 years ago

How does the frequency of infrared electromagnetic waves compare with the frequency of radio and microwaves?

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Answer:

Answer is B.

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An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?

Scorpion4ik [409]

Answer:

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Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

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3 years ago

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

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so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

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3 years ago